Question

Determine how many terms of the following convergent series must be summed to be sure that the remainder is less than \(10^{-4} .\) Although you do not need it, the exact value of the series is given in each case. $$\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{(2 k+1) !}$$

Short answer

Answer: At least 3 terms need to be added.

Step-by-step solution

Identify the n-th term of the series

The given series is $$\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{(2k+1)!}$$ From this, we can identify the n-th term as: $$a_n = \frac{(-1)^{n+1}}{(2n+1)!}$$

Find when the absolute value of the n-th term is less than the desired value

As the reminder of an alternating series is given by the absolute value of the n-th term, we need to find when \(|a_n|<10^{-4}\): $$\left|\frac{(-1)^{n+1}}{(2n+1)!}\right|<10^{-4}$$ Since \((-1)^{n+1} \le 1\), we need to find when: $$\frac{1}{(2n+1)!}<10^{-4}$$

Determine how many terms must be summed

We now need to find the smallest integer value for \(n\) that satisfies the inequality above. We can do that by checking n values sequentially: For \(n=1\): $$\frac{1}{(2(1)+1)!}=\frac{1}{6}<10^{-4}$$ \(\frac{1}{6}\) is not less than \(10^{-4}\), we need to try with the next value. For \(n=2\): $$\frac{1}{(2(2)+1)!}=\frac{1}{120}<10^{-4}$$ \(\frac{1}{120}\) is still not less than \(10^{-4}\), try with the next value. For \(n=3\): $$\frac{1}{(2(3)+1)!}=\frac{1}{5040}<10^{-4}$$ \(\frac{1}{5040}\) is less than \(10^{-4}\), so we found the smallest integer value we were looking for. Hence, we need to sum at least 3 terms of the series to be sure that the remainder is less than \(10^{-4}\).

Key concepts

Convergence of Series

In mathematics, understanding whether a series converges or diverges is crucial. A series converges if its terms approach a specific value as you sum more and more terms. Consider the series \(\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{(2k+1)!}\), which is an example of an alternating series, where the sign of the terms alternates between positive and negative.

For an alternating series to converge, it must meet two conditions:

• The absolute value of the series' terms \(a_n\) must decrease monotonically, meaning each term is smaller than the previous one.
• The limit of \(a_n\) as \(n\) approaches infinity must be zero: \(\lim_{n \to \infty} a_n = 0\).

The series given meets these requirements, indicating it converges. The presence of factorials \((2k+1)!\) results in terms that diminish rapidly, thereby ensuring convergence. By establishing the convergence of such a series, we can confidently discuss the behavior of its sums.

Remainder Estimation

Estimating the remainder of a series is an essential concept when determining how close our summation is to the series' total value. For alternating series, the remainder after \(n\) terms, denoted as \(R_n\), is the difference between the sum of the first \(n\) terms and the entire series.

A useful property of alternating series is that the absolute value of the remainder \(R_n\) is bounded by the absolute value of the first neglected term, \(|a_{n+1}|\). This makes finding the remainder straightforward:

• Simply find \(n\) such that \(|a_{n+1}| < \text{desired accuracy}\).

Hence, for the problem given, we know \(|a_n|\) should be less than \(10^{-4}\). By examining \(\frac{1}{(2n+1)!} \) for successive values of \(n\), we determine the smallest \(n\) that satisfies this inequality. This calculation forms the basis of estimating remainders in alternating series.

Factorials

In mathematics, factorials play a critical role, especially in series and combinatorics. The factorial function, denoted by \(n!\), is the product of all positive integers up to \(n\). Thus, \(n! = n \times (n-1) \times (n-2) \times \ldots \times 1\).

Factorials grow extremely fast. Even small values of \(n\) lead to relatively large factorial results.

• For example, \(5! = 120\) and \(7! = 5040\).
• This rapid increase in size greatly influences series, typically causing terms to become very small very quickly.

In our series \(\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{(2k+1)!}\), factorials are in the denominator. This makes the terms decrease swiftly, ensuring that as \(k\) increases, the value of each term becomes tiny, bolstering quick convergence of the series.

Understanding how factorials operate in different mathematical contexts is key to analyzing and estimating the behavior of series like the one presented.